OFFSET
1,1
COMMENTS
Subsequence of A212662.
a(1)=3 is the only prime term.
It is not known if this sequence is finite.
Every term in A212662 is a multiple of a term in this sequence (this could be considered its primitive sequence), and no term in this sequence divides another term of this sequence.
In general, we do not have (i+j)'=i'+j'; this is in contrast with the normal derivative, because the derivative of the sum of two functions is equal to the sum of the derivatives of the individual functions. The terms in this sequence are the ones for which there exist positive integers i and j with the properties illustrated above, thus building another common point between the arithmetic derivative and the normal derivative.
PROG
(PARI)
D(n)={x=factor(n); n*sum(i=1, matsize(x)[1], }{x[i, 2]/x[i, 1])}
Base(n)={for(i=1, n\2, if(D(n-i)+D(i)==D(n)&&gcd(i, n-i)==1, return(1)))}
Impure(n)={for(i=1, n\2, if(D(n-i)+D(i)==D(n)&&gcd(i, n-i)!=1, return(1)))}
IsTerm(n)={Base(n)&&!Impure(n)}
Bp(n)={for(i=2, n, if(IsTerm(i), print1(i, ", ")))}
Bp(1000)
(PARI) ard(n)=vecsum([n/f[1]*f[2]|f<-factor(n+!n)~]); \\ A003415
isok1(m) = for (k=1, m\2, if (ard(m-k)+ard(k) == ard(m), return(1))); \\ A212662
isok(m) = if (isok1(m), my(d=divisors(m)); for (k=1, #d, if((d[k]!=m) && isok1(d[k]), return(0))); return(1)); \\ Michel Marcus, Aug 28 2022
CROSSREFS
KEYWORD
nonn
AUTHOR
Giosuè Cavallo, Aug 14 2022
STATUS
approved